2025 AMC 10B Problem 4

Write each numeral in place-value form: $\underline{a}~\underline{b}$ in base seven is $7a + b$, and $\underline{b}~\underline{a}$ in base nine is $9b + a$. Setting them equal,
$$7a + b = 9b + a \quad\Longrightarrow\quad 6a = 8b \quad\Longrightarrow\quad 3a = 4b.$$

Now use the digit constraints. Both $a$ and $b$ appear as base-seven digits, so $0 \le a, b \le 6$, and both are leading digits of a two-digit numeral, so $a, b \ge 1$.

From $3a = 4b$ and $\gcd(3,4) = 1$, the factor $4$ must divide $a$. The only multiple of $4$ in the range $1 \le a \le 6$ is $a = 4$, which gives $b = 3$.

Check: $43_7 = 7 \cdot 4 + 3 = 31$ and $34_9 = 9 \cdot 3 + 4 = 31$. Hence $a + b = 7$.

The answer is $\boxed{\textbf{(A)}\ 7}$.

Converting both numerals to base ten immediately is the whole method: a statement about digits becomes a linear equation, and the only extra information needed is the range each digit may occupy. The coprimality step ($4 \mid a$) is what turns one equation in two unknowns into a finite check, and it is faster than listing all pairs. The same reduction handles any "numeral in base $m$ equals its reversal in base $n$" problem: the equation is $(m-1)a = (n-1)b$ after cancellation.

• Solving 3a = 4b without the base-seven bound and accepting a = 8, b = 6, which gives a + b = 14, choice (E).
• Reading $\underline{a}~\underline{b}$ as the product $ab$ rather than as a two-digit numeral.
• Forgetting that $b$ must also be a valid base-seven digit, which would wrongly allow $b$ up to $8$.